3.10.88 \(\int \frac {e (15+15 x)+e (45+15 x) \log (3+x)}{3+x} \, dx\)

Optimal. Leaf size=12 \[ 5 e (3+3 x) \log (3+x) \]

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Rubi [A]  time = 0.10, antiderivative size = 18, normalized size of antiderivative = 1.50, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6688, 12, 6742, 43, 2389, 2295} \begin {gather*} 15 e (x+3) \log (x+3)-30 e \log (x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E*(15 + 15*x) + E*(45 + 15*x)*Log[3 + x])/(3 + x),x]

[Out]

-30*E*Log[3 + x] + 15*E*(3 + x)*Log[3 + x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 e (1+x+(3+x) \log (3+x))}{3+x} \, dx\\ &=(15 e) \int \frac {1+x+(3+x) \log (3+x)}{3+x} \, dx\\ &=(15 e) \int \left (\frac {1+x}{3+x}+\log (3+x)\right ) \, dx\\ &=(15 e) \int \frac {1+x}{3+x} \, dx+(15 e) \int \log (3+x) \, dx\\ &=(15 e) \int \left (1-\frac {2}{3+x}\right ) \, dx+(15 e) \operatorname {Subst}(\int \log (x) \, dx,x,3+x)\\ &=-30 e \log (3+x)+15 e (3+x) \log (3+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.17 \begin {gather*} 15 e (\log (3+x)+x \log (3+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E*(15 + 15*x) + E*(45 + 15*x)*Log[3 + x])/(3 + x),x]

[Out]

15*E*(Log[3 + x] + x*Log[3 + x])

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fricas [A]  time = 0.59, size = 11, normalized size = 0.92 \begin {gather*} 15 \, {\left (x + 1\right )} e \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x+45)*exp(1)*log(3+x)+(15*x+15)*exp(1))/(3+x),x, algorithm="fricas")

[Out]

15*(x + 1)*e*log(x + 3)

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giac [A]  time = 0.27, size = 18, normalized size = 1.50 \begin {gather*} 15 \, x e \log \left (x + 3\right ) + 15 \, e \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x+45)*exp(1)*log(3+x)+(15*x+15)*exp(1))/(3+x),x, algorithm="giac")

[Out]

15*x*e*log(x + 3) + 15*e*log(x + 3)

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maple [A]  time = 0.16, size = 19, normalized size = 1.58

method	result	size
norman	\(15 \,{\mathrm e} \ln \left (3+x \right )+15 x \,{\mathrm e} \ln \left (3+x \right )\)	\(19\)
risch	\(15 \,{\mathrm e} \ln \left (3+x \right )+15 x \,{\mathrm e} \ln \left (3+x \right )\)	\(19\)
derivativedivides	\(15 \,{\mathrm e} \left (\left (3+x \right ) \ln \left (3+x \right )-3-x \right )+15 \left (3+x \right ) {\mathrm e}-30 \,{\mathrm e} \ln \left (3+x \right )\)	\(34\)
default	\(15 \,{\mathrm e} \left (\left (3+x \right ) \ln \left (3+x \right )-3-x \right )+15 \left (3+x \right ) {\mathrm e}-30 \,{\mathrm e} \ln \left (3+x \right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*x+45)*exp(1)*ln(3+x)+(15*x+15)*exp(1))/(3+x),x,method=_RETURNVERBOSE)

[Out]

15*exp(1)*ln(3+x)+15*x*exp(1)*ln(3+x)

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maxima [B]  time = 0.34, size = 69, normalized size = 5.75 \begin {gather*} 15 \, {\left (x - 3 \, \log \left (x + 3\right )\right )} e \log \left (x + 3\right ) + \frac {45}{2} \, e \log \left (x + 3\right )^{2} + \frac {15}{2} \, {\left (3 \, \log \left (x + 3\right )^{2} - 2 \, x + 6 \, \log \left (x + 3\right )\right )} e + 15 \, {\left (x - 3 \, \log \left (x + 3\right )\right )} e + 15 \, e \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x+45)*exp(1)*log(3+x)+(15*x+15)*exp(1))/(3+x),x, algorithm="maxima")

[Out]

15*(x - 3*log(x + 3))*e*log(x + 3) + 45/2*e*log(x + 3)^2 + 15/2*(3*log(x + 3)^2 - 2*x + 6*log(x + 3))*e + 15*(
x - 3*log(x + 3))*e + 15*e*log(x + 3)

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mupad [B]  time = 0.83, size = 11, normalized size = 0.92 \begin {gather*} 15\,\ln \left (x+3\right )\,\mathrm {e}\,\left (x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1)*(15*x + 15) + log(x + 3)*exp(1)*(15*x + 45))/(x + 3),x)

[Out]

15*log(x + 3)*exp(1)*(x + 1)

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sympy [A]  time = 0.12, size = 20, normalized size = 1.67 \begin {gather*} 15 e x \log {\left (x + 3 \right )} + 15 e \log {\left (x + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x+45)*exp(1)*ln(3+x)+(15*x+15)*exp(1))/(3+x),x)

[Out]

15*E*x*log(x + 3) + 15*E*log(x + 3)

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